Let the latus rectum of the parabola $$y^2 = 4x$$ be the common chord to the circles $$C_1$$ and $$C_2$$ each of them having radius $$2\sqrt{5}$$. Then, the distance between the centres of the circles $$C_1$$ and $$C_2$$ is:

We have the parabola $$y^{2}=4x$$. For any parabola of the form $$y^{2}=4ax$$, the focus is $$(a,0)$$ and the latus-rectum is the line $$x=a$$ whose length is $$4a$$.

Comparing $$y^{2}=4x$$ with $$y^{2}=4ax$$ gives $$a=1$$. Therefore

$$\text{Equation of latus rectum}=x=a=1,$$

$$\text{Length of latus rectum}=4a=4.$$

The points where this line meets the parabola are obtained by substituting $$x=1$$ in $$y^{2}=4x$$:

$$y^{2}=4(1)=4 \;\;\Longrightarrow\;\; y=\pm2.$$

Hence the latus-rectum is the segment joining $$(1,2)$$ and $$(1,-2)$$, so its length is indeed

$$\sqrt{(1-1)^{2}+(2-(-2))^{2}}=\sqrt{0+4^{2}}=4.$$

This segment is given to be the common chord of two equal circles $$C_{1}$$ and $$C_{2}$$, each having radius

$$R=2\sqrt{5}.$$

Let $$O_{1}$$ and $$O_{2}$$ be the centres of $$C_{1}$$ and $$C_{2}$$ respectively, and let $$d$$ be the (perpendicular) distance from either centre to the chord. For a circle, the length $$2\ell$$ of a chord at distance $$d$$ from the centre satisfies

$$\ell^{2}=R^{2}-d^{2}.$$

Here the full chord length is $$4$$, so the half-length is $$\ell=2$$. Substituting $$\ell=2$$ and $$R=2\sqrt{5}$$ gives

$$2^{2}=R^{2}-d^{2} \;\;\Longrightarrow\;\; 4=(2\sqrt{5})^{2}-d^{2}$$

$$4=20-d^{2}$$

$$d^{2}=20-4=16 \;\;\Longrightarrow\;\; d=4.$$

The latus-rectum is the vertical line $$x=1$$, so the perpendicular distance is measured horizontally. Hence the $$x$$-coordinate of each centre is $$4$$ units away from $$x=1$$:

$$x=1+4=5 \quad\text{or}\quad x=1-4=-3.$$

The chord’s midpoint is $$(1,0)$$, lying on the $$x$$-axis, and the circles are symmetric about this axis, so the centres must have $$y=0$$. Therefore

$$O_{1}=(5,0), \qquad O_{2}=(-3,0).$$

The distance between the centres is then

$$|5-(-3)|=\;8.$$

Hence, the correct answer is Option B.